10-4t-5t^2=0

Solution for 10-4t-5t^2=0 equation:

10-4t-5t^2=0

a = -5; b = -4; c = +10;
Δ = b2-4ac
Δ = -42-4·(-5)·10
Δ = 216
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-6\sqrt{6}}{2*-5}=\frac{4-6\sqrt{6}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+6\sqrt{6}}{2*-5}=\frac{4+6\sqrt{6}}{-10}
$